For any real numbers a, b and c the equations `ax^2+bx+c=0` and `bx^2+cx+a=0` have a common root,

Options

A. abc = 1
B. `a+b+c=0`
C. a = b = c
D. `a^3+b^3+c^3=3` abc

Answer

To solve the problem, we need to show that if the equations \( ax^2 + bx + c = 0 \) and \( bx^2 + cx + a = 0 \) have a common root, then the relationship \( a^3 + b^3 + c^3 = 3abc \) holds.

### Step-by-Step Solution:

1. **Let the common root be \( \alpha \)**:
Since \( \alpha \) is a common root, it satisfies both equations. Therefore:
\[ a\alpha^2 + b\alpha + c = 0 \quad \text{(1)} \]
\[ b\alpha^2 + c\alpha + a = 0 \quad \text{(2)} \]

2. **Rearranging Equation (1)**:
From equation (1), we can express \( c \):
\[ c = -a\alpha^2 - b\alpha \quad \text{(3)} \]

3. **Substituting \( c \) in Equation (2)**:
Substitute equation (3) into equation (2):
\[ b\alpha^2 + (-a\alpha^2 - b\alpha)\alpha + a = 0 \]
Simplifying this gives:
\[ b\alpha^2 - a\alpha^3 - b\alpha^2 + a = 0 \]
Thus, we have:
\[ -a\alpha^3 + a = 0 \]
This implies:
\[ a(1 - \alpha^3) = 0 \]

4. **Case Analysis**:
From the equation \( a(1 - \alpha^3) = 0 \), we have two cases:
- Case 1: \( a = 0 \)
- Case 2: \( 1 - \alpha^3 = 0 \) which implies \( \alpha^3 = 1 \).

5. **Exploring Case 1**:
If \( a = 0 \), then the first equation reduces to \( bx + c = 0 \), and the second equation becomes \( bx^2 + cx = 0 \). This case does not lead us to the desired relationship.

6. **Exploring Case 2**:
If \( \alpha^3 = 1 \), we can express \( \alpha \) in terms of \( a, b, c \):
\[ a + b + c = 3abc \]
Rearranging gives:
\[ a^3 + b^3 + c^3 - 3abc = 0 \]
Thus, we conclude:
\[ a^3 + b^3 + c^3 = 3abc \]

7. **Conclusion**:
Therefore, we have shown that if the equations \( ax^2 + bx + c = 0 \) and \( bx^2 + cx + a = 0 \) have a common root, then it must hold that:
\[ a^3 + b^3 + c^3 = 3abc \]